Hi and welcome to Module 29 of Two-Dimensional Dynamics. Today's learning outcome is to solve an engineering problem for the kinetics of rigid body translating in two-dimensional planar motion. So this is a theory we developed last time, we came up with the equations, and we're gonna now apply this theory to this problem. And I'm gonna change the nomenclature here slightly so that we don't get confused. I'm gonna call the force F, and then to be consistent with my theory, I'm gonna call this point of contact P. And so I want to find the angle for this bar for which the bar will translate to the right with a constant in the angular, or excuse me a constant acceleration. And then I also want to find what force is required to maintain that translation. So let's look at a demo. I always like to look at the real world or physical significance of these problems. So here is my bar. I've got a surface with the coefficient of friction mu. I've got a force f down here and I wanna push to the right where it stays at a constant angle and a constant acceleration and I'm not quite doing it, but I'm getting close. And so we want it to go along with a constant acceleration or translation. And so my first question to you is, how do we approach this problem, and how are you gonna start? And what you should do, this is often the most difficult part of the problem, to set it up. And what you wanna do here is, I like to use these graphical tools, like I said, in statics you always wanna draw a free body diagram. Let's draw a free body diagram, but now we have kinetic or motion vectors in the right hand side of our equation, so I'm also gonna use a kinetic diagram. And so let's go ahead and have you begin by doing the free body diagram, and come on back and see how you did. And so here's the free body diagram, you can see if you got that correct. Let's now proceed with the kinetic diagram. So here is my body of interest, the bar. Here is the mass center, C. And I'm gonna put my kinetic vectors or motion vectors at the mass center, ma is the vector. I'll break it into two orthogonal components. I'll use the x and y directions, so I'm gonna have an max, and an may. Can we simplify that kinetic diagram at all, though? And what you should say is, yes. If this thing is translating, may should be equal to zero. And so let's go back and look at my demonstration, and again, if this is translating, then the acceleration of the mass center is only in the horizontal direction, there's no acceleration in the vertical direction, and so that's a simplification. So we'll just go ahead and call max, we'll just call it ma, since there's only one component of acceleration. All right, let's pick up from there. And here is my free body and kinetic diagram, free body diagram and kinetic diagram, what do I do next? And so what you should says is I'm gonna go ahead and apply my equations, I can either sum moments about P or C, P being any arbitrary point. I'll actually pick point P. That's sort of a smart point, if you will, because It gets rid of three of my unknowns. All three of these forces have a line of action through point P, so they're not gonna cause a moment about point P. So let's write the equations, sum of the moments about P on the free body diagram equals sum of the moments about P on the kinetic diagram. I'll choose counterclockwise positive for assembling my equation. And so if I sum moments about P on my free body diagram, the only force that's gonna cause a moment is the weight force mg. It'll tend to cause a clockwise rotation, which is opposite my positive sign convention, so it's gonna be minus the force mg times it's moment arm, which is gonna be L over 2 cos theta equals, that's the only force causing, with a tendency to cause a rotation or moment about P on the free body diagram. Over here on the kinetic diagram, I've got my ma vector, or effective force. It's gonna tend to cause a clockwise rotation. And so that's again, negative in accordance with my sign convention. Its moment arm will be L/2 sin theta. Then I can simplify this equation by cancelling the m's. Canceling L over 2, and I get g over a = sin theta over cos theta, but we all know that sin theta over cos theta is tan theta, and so our result is that theta = tan inverse( g over a). And I've now solved one portion of the problem. I have now found the angle for which the bar translates to the right with a constant acceleration. Whatever that acceleration is, if I divide that into my acceleration due to gravity and take the inverse tangent, that'll tell me what my angle is. And so let's continue on. There's the results so far. Next I wanna find the force F required to maintain that translation. How would I do that? And what you should say is well, let's go ahead and do another equation for forces in the x direction, since we want to find that force F. And so I've got the sum of the forces in the F direction on my free body diagram equals the sum of the forces in the x direction on my kinetic diagram. I'll choose to the right positive. And so I get F- mu N = ma, or F = ma plus mu N. All right, let's use that result. So so far we've got theta, and we've got the force, but we need to put force, we have to find out what the normal force is, and so to do that we'll go ahead and sum forces in the y direction for both my free body diagram and kinetic diagram, the sum of the forces and the y of the free body diagram equals the sum of the forces in the y of the kinetic diagram. I'll choose up, positive. And so I get N- mg, N is up, positive, mg is down, negative, equals no kinetic vectors in the y direction, 0. So N = mg. We can then go ahead and substitute that in up here. And I arrive at my force equals ma + mu times mg. And I can factor out the m, so I get F = m (mu g + a), and so the force required will be whatever the mass of the bar is times the quantity of the coefficient of friction times acceleration due to gravity plus the constant acceleration, and so for the higher acceleration I'm gonna need a larger force. For larger mass I'm gonna need a larger force. All makes physical sense, okay? So here's our result, we have our angle theta and the force required to have translation to the right. We found that result by summing moments about an arbitrary point P which was the point of contact on our body. Let's just double-check to make sure that the sum of the moments about the mass center equals zero. So we're gonna do a check here. So I've got the sum of the moments about C on the free body diagram equals the sum of the moments about C on the kinetic diagram. I'll choose counterclockwise for assembling my equation as my sign convention. And so The F force will tend to cause a counterclockwise rotation, about point C. So that's gonna be positive, in accordance with my sign convention. Its moment arm here is going to be L over 2 sin theta. Now I have the N force, it's gonna tend to cause a clockwise rotation about C, so that's gonna be negative in accordance with my sin convention. So I got -N times its moment arm, and that moment arm is going to be, here's the line of action, here's the point, distance between the two is L over 2 cos theta. And then I've got my mu N force about point C. It's also gonna tend to cause a clockwise rotation, which is negative in accordance with my sign convention. So I have mu sub N times its moment arm, which again, is L over 2 sin theta, and that equals, since the effective force or the motion vector on my kinetic diagram, the line of action goes through point C, there's no moment there, so it's equal to zero. The moment arm is zero. And so I can now collect terms. Let's collect this first and third term here so I get F, In fact, before I even do that, let's cancel out the L over 2. That'll make it a little easier here. And so I've got (F- mu n) sin theta. And that takes care of this term and this term. =, I'll move this term to the other side of the equation, and I get N cos theta. But F- mu N, if I put F- mu N, that's equal to ma, and N we said was equal to mg. And so I've got, let's see, this should be ma, not mg, so I've got ma sin theta = mg cos theta, the m's cancel, and I get tan theta, if I divide cosine, is equal to g over a or, excuse me, that's going to be theta is the inverse tangent of g over a and so sum of the moments about zero does indeed check. And so that's a good translation problem. I'd like you to try one on your own. Here's a worksheet for you to do, and I've got the solution in the handouts. And we'll see you next time.